Q 11-02-110JEE MainJEE Main 2021 (25 Jul, Shift 2)Medium
The relation between time $t$ and distance $x$ for a moving body is given as $t = mx^2 + nx$, where $m$ and $n$ are constants. The retardation of the motion is: (When $v$ stands for velocity)
Answer: (A) $2mv^3$
Differentiate with respect to $t$: $1 = (2mx + n)v$, so $\dfrac{1}{v} = 2mx + n$.
Differentiate again: $-\dfrac{1}{v^2}\dfrac{dv}{dt} = 2m\dfrac{dx}{dt} = 2mv$
$$a = -2mv^3$$
The retardation is $2mv^3$.
Solution by Sreeraj P, M.Sc Physics